What is the oxidation state of the potassium permanganate?

1 Answer
Feb 16, 2015

I assume the question refers to the oxidation number of manganese, or #Mn#, in potassium permanganate, or #KMnO_4#.

If you're familiar with the rules for calculating oxidation numbers (more here: http://socratic.org/questions/how-do-you-calculate-oxidation-numbers), you'll notice immediately that each oxygen atom has an O.N. of -2 and that the potassium atom has an O.N. of +1.

Since #KMnO_4# is a neutral compound, the sum of the O.N. calculated for each atom must be equal to zero. This implies that

#ON_(K) + ON_(Mn) + 4ON_(O) = 0#, or

#(+1) + ON_(Mn) + 4 * (-2) = 0 => ON_(Mn) = 0 -1 + 8#

#ON_(Mn) = +7#

As a side note, potassium permanganate is often used in experiments that show the color changes that accompany the reduction of manganese from +7 to +6, and from +6 to +4.

This is done by reacting #KMnO_4# with a solution containing sodium hydroxide and sucrose (regular table sugar). The rezulting solution will go from being purple (ON +7), to being green (ON +6) and finally to being yellow (ON +4).

Here's a video on such an experiment: